Let $\bf{A_{11}}$ $\in$ $\mathbb{F}^{pxp}$, $\bf{A_{12}}$ $\in$ $\mathbb{F}^{pxq}$, and $\bf{A_{21}}$ $\in$ $\mathbb{F}^{qxp}$. Show that if $\bf{A_{11}}$ is invertible, then
a) The block matrix [$\bf{A_{11}}$ $\bf{A_{12}}$] is right invertible
b) The block matrix $[\bf{A_{11}} \bf{A_{21}}]^T$ is left invertible
Hint: You have to construct the right and left inverses as block matrices.
I don't really have an idea of how to tackle this problem, but I think that if $\bf{A_{11}}$ is invertible, then C$\bf{A_{11}}$ = $\bf{I}$, for the left inverse and $\bf{A_{11}}$ B=$\bf{I}$ for the right inverse. Nonetheless, I don't quite see how this would transfer to the block matrices. I also don't understand how the hint would apply. How could I construct the right and left inverses of the block matrices. Any help would be appreciated.
Let $B = \begin{bmatrix}X\\Y\end{bmatrix}$ be a right inverse of $A = \begin{bmatrix}A_{11} & A_{12}\end{bmatrix}$.
Then \begin{align*} AB = I \implies \\A_{11}X + A_{12}Y = I \implies\\ X = A_{11}^{-1}(I- A_{12}Y). \end{align*}
So you can choose any matrix of suitable order as $Y$, and define $X$ accordingly to get a right inverse of $A$.
The left invertibility of the other matrix follows immediately since its transpose is a right invertible matrix by what has been proved already.